Question

A rectangular channel having a bed slope of 0.0001, width 3.0 m, and Manning's coefficient 'n' 0.015, carries a discharge of $1.0m^3/s$. Given that the normal depth of flow ranges between 0.76 m and 0.8 m. The minimum width of a throat (in m) that is possible at a given section, while ensuring that the prevailing normal depth is not exceeded along the reach upstream of the contraction, is approximately equal to (assume negligible losses)

• A. 0.64
• B. 1.04
• C. 1.24
• D. 0.84

Answer 1

The correct option is D 0.84

$n=0.015$
$Q=1m^3/s$
Normal depth of flow between 0.76 m to 0.8 m.
If prevailing normal depth of flow is not exceeded, there must not be chocking of the section or there must be just chocking.
Thus the width of the section should be such that for the prevailing specific energy there should be critical flow at the contracted section
i.e. $\frac{3}{2}{\left(\frac{q^2}{g}\right)}^{1/3}=E_C=E_{initial}$

$\Rightarrow \frac{3}{2}{\left[\frac{{\left(\frac{Q}{B_{min}}\right)}^2}{g}\right]}^{1/3}=E_{initial}$

Let us now calculate $E_{initial}$

$Q=\frac{1}{2}A{R}^{2/3}{S}_0^{1/2}$

$\Rightarrow 1=\frac{1}{0.015}\left(3y\right){\left(\frac{3y}{3+2y}\right)}^{2/3}(0.0001{)}^{1/2}$

$\Rightarrow y=0.78m$

$\Rightarrow E_{initial}=y+\frac{q^2}{2g{y}^2}$

$=0.78+\frac{{\left(\frac{1}{3}\right)}^2}{2\times 9.81\times (0.78{)}^2}=0.7893m$

$\Rightarrow \frac{3}{2}{\left[\frac{{\left(\frac{Q}{B_{min}}\right)}^2}{g}\right]}^{1/3}=0.7893$

$\Rightarrow \frac{3}{2}\frac{(1{)}^{2/3}}{g^{1/3}(B_{min}{)}^{2/3}}=0.7893$

$B_{min}=0.836m$